Question

what is (cosx/sinx)/sinxcosx simplified

Answer 1

\[\frac{ \cos(x) }{ \sin(x) } \times \frac{ \sin(x)\cos(x) }{ 1 }\]

Answer 2

\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x}\]Woops I dunno if you're looking at it correctly abbot :O Only the top one is a fraction.

Answer 3

\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \left(\frac{\cos x}{\sin x}\right)\frac{1}{\sin x \cos x}\]

Answer 4

ya but what about multiplying by the recipricol.... cause i am seriously confused.

Answer 5

\[\large \left(\frac{\cancel{\cos x}}{\sin x}\right)\frac{1}{\sin x \cancel{\cos x}}\]

Answer 6

Ok here's the thing with the reciprocal :) lemme explain.

Answer 7

ok sounds good

Answer 8

The bottom fraction is actually this,\[\large \frac{\sin x \cos x}{1}\]

So if you wanted to write it as a division of fractions, you could write it like this,\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\left(\dfrac{\sin x \cos x}{1}\right)}\]And then from here, we could multiply by the reciprocal of the bottom fraction.

Answer 9

aha!!!!! now it's making sense!!!

Answer 10

When we started this problem, we weren't dividing fractions. Only the top one was a fraction.
Maybe that's why there was a little confusion :)

Answer 11

i think so...this is the only one that i've been confused over so far... thank you:)

Answer 12

i got the right answer. thanks a billion

Answer 13

yay!